AP Physics 1 : Specific Forces

Study concepts, example questions & explanations for AP Physics 1

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Example Questions

Example Question #24 : Force Of Friction

How will doubling the coefficient of friction of rubber tires change the stopping distance of a car?

Possible Answers:

Impossible to determine

Cut the stopping distance in half

None of these

Quadruple the stopping distance

Double the stopping distance

Correct answer:

Cut the stopping distance in half

Explanation:

Since the car is stopped at the end, 

Combining equations:

Solving for 

From this, it can be seen that doubling the coefficient of friction would cut the stopping distance in half.

Example Question #731 : Ap Physics 1

A box is shoved and is sliding on a concrete floor. It has a mass of  and in  slows from  to . Determine the coefficient of friction.

Possible Answers:

None of these

Correct answer:

Explanation:

Work is due to friction:

Solving for 

Plugging in values:

Example Question #731 : Ap Physics 1

A box is shoved and is sliding on a concrete floor. It has a mass of  and in  slows from  to . Determine the coefficient of friction.

Possible Answers:

None of these

Correct answer:

Explanation:

Work is due to friction:

Solving for 

Plugging in values:

Example Question #701 : Newtonian Mechanics

A  block is held at rest at the top of slope inclined at . When the block is let go it slides down the slope, experiencing friction along the way. If the acceleration of the block down the ramp is , what is the coefficient of kinetic friction between the block and ramp?

Possible Answers:

Correct answer:

Explanation:

Img 20160212 183040582Because the block is traveling on an inclined ramp and experiences friction, the two main forces at work here are a component of gravitational force and kinetic frictional force.

Img 20160212 183048473 The total force acting on the block can be written as , where  is the component of gravitational force in the direction of motion, and  is the kinetic frictional force.

We know that forces can be expressed as , and kinetic friction can be expressed as . Hence our total force can be expressed as . The normal force can be further expressed as 

The component of gravity parallel to the ramp is given by , while the component of gravity perpendicular to the ramp is given by .

Our total force can hence be expressed as .

Thus, 

and 

Example Question #33 : Force Of Friction

Jennifer has a mass of  and is riding a rocket powered sled of mass . The rocket produces a thrust of  and she accelerates at a rate of . Determine the coefficient of friction between the sled and the snow underneath.

Possible Answers:

Correct answer:

Explanation:

Solving for the coefficient of friction:

Example Question #34 : Force Of Friction

Consider the following scenario:

Sledder

A sledder of mass  is at the stop of a sledding hill at height  with a slope of angle .

The hill has a height of  and a slope . If a 75-kilogram sledder is initially at rest at the top of the hill and reaches the bottom of the hill with a final velocity , what is the average frictional force applied to the sledder? Neglect air resistance and any other frictional forces.

Possible Answers:

Correct answer:

Explanation:

We can use the expression for conservation of energy to solve this problem:

The problem statement tells us that the sledder is initially at rest, so we can eliminate initial kinetic energy. Also, if we assume that the bottom of the hill has a height of 0, we can eliminate final potential energy to get:

  (1)

From here we need to determine what is going to contribute to work. The only extra piece of information we have in this problem is that there is an average frictional force. Therefore that will be the only source of work. Let's began expanding each term, going from left to right:

  (2)

Note that we didn't mark the height as an initial height because we assumed that the bottom of the hill has a height of 0.

Where d is the length of the hill. We can calculate this distance using the height and angle of the hill.

Rearranging for distance:

Substituting this back into our expression for work, we get:

  (3)

Now our final term:

    (4)

Now substituting expression 2, 3, and 4 into expression 1, we get:

The reason we are subtracting friction is because it is removing energy from the system. Another way we could have written our initial expression would be to have work on the final state and then friction would be positive.

Rearranging for the frictional force:

We have all of these values, so time to plug and chug:

Example Question #1 : Spring Force

A horizontal spring with a spring constant of  is attached to a frictionless surface. A block of mass 2kg is attached to the end of the spring. A man spends 20J of energy to compress the spring. How far from equilibrium is the block?

Possible Answers:

Correct answer:

Explanation:

You simply need to know the formula for the potential energy stored in a spring to solve this problem. The formula is:

where k is the spring constant, and x is the distance from equilibrium

We can rearrange this to get:

Plugging in our values, we get:

Example Question #1 : Spring Force

Consider the following system:

 Img_2103

Both springs have a constant of  and the block is motionless. If the bottom spring is compressed  past its equilibrium and the block has a mass of , how far is the top spring stretched past its equilibrium?

Possible Answers:

Correct answer:

Explanation:

Since the block is motionless, we know that our forces will cancel out:

There are three forces in play: one from each spring, as well as the force of gravity. If we assume that forces pointing up are positive, we can write:

Plugging in expressions for each spring force, we get:

Rearring for the displacement of the top spring, we get:

Example Question #1 : Spring Force

Consider the following system:

Img_2103

Both springs have a constant of  and the block is motionless. If the top spring is stretched  past its point of equilibrium and the bottom spring is compressed  past its point of equilibrium, what is the mass of the block?

Possible Answers:

Correct answer:

Explanation:

Since the block is motionless, we can assume that the forces cancel out:

If we designate any forces pointing downward as positive, we can write:

Inserting expressions for each force, we get:

Rearranging for mass, we get:

Example Question #2 : Spring Force

A spring attached to the ceiling has a block of mass  attached to the other end. On Earth, the displacement of the spring when at equilibrium is . What is the displacement of the spring at equilibirum when this system is on a planet that has a gravitational acceleration of

Possible Answers:

Correct answer:

Explanation:

There are two ways to solve this problem: the first involves calculating the spring constant and the second does not. We'll go through both methods.

Calculating Spring Constant

We can use the expression for the force of a spring:

At equilibrium, the force of the spring equals the force of gravity:

Rearranging for the spring constant and plugging in values, we get:

Now, apply this equation when the spring is on a different planet:

Rearranging for displacement and plugging in values, we get:

Without Calculating Spring Constant

We can write the force equation for each scenario. Let the subscript 1 denote Earth, and 2 denote the other planet:

Using these equations, we can set up a proportion:

Rearranging for the displacement of scenario 2 and plugging in values:

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